A random sample of the closing stock prices in dollars for a company in a recent

Question

A random sample of the closing stock prices in dollars for a company in a recent year is listed below. Assume that sigma is $2.28. Construct the 90% and 99% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. The 90% confidence interval is (Round to two decimal places as needed.) The 99% confidence interval is (Round to two decimal places as needed.) Which statement below interprets the results correctly? There is 90% confidence that the mean closing stock price is in the 90% confidence interval and 99% confidence that the mean closing stock price is in the 99% confidence interval. The 90% confidence interval contains the mean closing stock price 90% of the time and the 99% confidence interval contains the mean closing stock price 99% of the time. The probability that the mean closing stock price is in the 90% confidence interval is about 90% and the probability that the mean closing stock price is in

Explanation / Answer

a.
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=18.5488
Standard deviation( sd1 )=2.3432
Sample Size(n1)=16
Mean(x2)=6.333
Standard deviation( sd2 )=1.52753
Sample Size(n2)=30
CI = [ ( 18.5488-6.333) ±t a/2 * Sqrt( 5.49058624/16+2.3333479009/30)]
= [ (12.2158) ± t a/2 * Sqrt( 0.4209) ]
= [ (12.2158) ± 1.753 * Sqrt( 0.4209) ]
= [11.0785 , 13.3531] ~ [11.08 , 13.35]

b.
CI = [ ( 18.5488-6.333) ±t a/2 * Sqrt( 5.49058624/16+2.3333479009/30)]
= [ (12.22) ± t a/2 * Sqrt( 0.42) ]
= [ (12.22) ± 2.947 * Sqrt( 0.42) ]
= [10.3 , 14.13]

c.
There is 90% confidence that the mean dosing stock price is ¡n the 90% confidence interval and 99% confidence that the mean dosing stock price ¡s in the 99% confidence

d.
99% is wider

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